Problem Set 5 ECON 687 Due in class Theorem 1. If Assumption 2 in Handout 4 holds, then∥∥∥∥∥ 1n n∑ i=1 Xiεi ∥∥∥∥∥ ∞ = Op (√ log(p) n ) , where p is the dimension of X. Proof. For any ε > 0, we can choose M ′ sufficiently large such that P (∥∥∥∥∥ 1n n∑ i=1 Xiεi ∥∥∥∥∥ ∞ ≥M ′ √ log(p) n ) ≤P (∥∥∥∥∥ 1n n∑ i=1 Xiεi ∥∥∥∥∥ ∞ ≥M ′ √ log(p) n , max k∈[p],i∈[n] |Xi,kεi| ≤Mn1/q+a ) + P ( max k∈[p],i∈[n] |Xi,kεi| ≥Mn1/q+a ) ≤P ( max k∈[p] ∣∣∣∣∣ 1n n∑ i=1 Xi,kεi1{|Xi,kεi| ≤Mn1/q+a} ∣∣∣∣∣ ≥M ′ √ log(p) n ) + P ( max i∈[n] ζn|εi| ≥Mn1/q+a ) ≤P ( max k∈[p] ∣∣∣∣∣ 1n n∑ i=1 Xi,kεi1{|Xi,kεi| ≤Mn1/q+a} − E(Xi,kεi1{|Xi,kεi| ≤Mn1/q+a}) ∣∣∣∣∣ ≥ M ′2 √ log(p) n ) + P ( max k∈[p] |E(Xi,kεi1{|Xi,kεi| ≤Mn1/q+a})| ≥ M ′ 2 √ log(p) n ) + P ( max i∈[n] ζn|εi| ≥Mn1/q+a ) ≤ ∑ k∈[p] P (∣∣∣∣∣ n∑ i=1 Xi,kεi1{|Xi,kεi| ≤Mn1/q+a} − E(Xi,kεi1{|Xi,kεi| ≤Mn1/q+a}) ∣∣∣∣∣ ≥ M ′2 √log(p)n ) + ( ζn Mna )q ≤ ∑ k∈[p] exp ( − (M ′)2 log(p)n/8∑ i∈[n] EX2i,kε2i +M ′ √ log(p)nMn1/q+a/6 ) + ( C M )q . ≤p exp ( − (M ′)2 log(p)n/4 nC +M ′ √ log(p)nM(n)1/q+a/6 ) + ε ≤p exp (−(M ′/2)2 log(p)/(8C))+ ε =p1−(M ′)2/(4C) + ε ≤ 2ε. 1. Explain the reason for each inequality in the above proof. Hint: (a) second inequality: check out the moment inequality, (b) fourth inequality: moment inequality or Markov inequality and EXi,kεi = 0. (c) fifth inequality: Bernstein’s inequality. 1 2. Suppose Z1, · · · , Zp are p standard normal random variables. (a) Show P(Zi > x) ≤ 1x exp −x2/2 √ 2pi for x > 0. Hint: P(Zi > x) = ∫∞ x exp−t 2/2 √ 2pi dt ≤ ∫∞ x t x exp−t 2/2 √ 2pi dt. (b) Show P(|Zi| > x) ≤ 2 exp −x2/2 √ 2pi for x > 1. (c) Show maxk∈[p] |Zk| = Op( √ log(p)). 2
